Find the remainder when $15^{23}+23^{23}$ is divided by 19.

The answer is 0.

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First note that $15^{23} + 23^{23} = (19 - 4)^{23} + (19 + 4)^{23}$ .

Now when we expand these binomials, a factor of $19$ will appear in every term of the expansions except the last of each one, so we end up with

$15^{23} + 23^{23} \equiv (-4)^{23} + 4^{23} \pmod{19} = \boxed{0}$ .